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    $az^n+z+1$ has a root in $|z|\le 2$

    P-MICNK

    Area
    Complex Analysis
    Topics
    Polynomials, Rouché, Zeros
    Status
    draft

    Let \(a\in \CC\) and \(n\geq 2\). Show that the following polynomial has at least one root in \(\abs{z} \leq 2\): \[ f(z) = az^n + z + 1 .\]

    Solution

    First suppose \[ |a|<2^{-n}. \] On \(|z|=2\), \[ |az^n|=|a|2^n<1\le |z+1|. \] Rouché’s theorem therefore shows that \(az^n+z+1\) and \(z+1\) have the same number of zeros in \(|z|<2\), namely one.

    Now suppose \(|a|\ge2^{-n}\). Then \(a\neq0\). Let \(z_1,\ldots,z_n\) be the roots of \(f\), counted with multiplicity. Vieta’s formula gives \[ \left|\prod_{k=1}^n z_k\right|={1\over|a|}\le2^n. \] If every root satisfied \(|z_k|>2\), their product would have modulus strictly greater than \(2^n\), a contradiction. Hence at least one root lies in \(|z|\le2\).

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